Question

An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.

The final velocity of the electrons is _____.
The mass of the electron is 9.1x10^(-31) kg and its charge is 1.6x10^(-19) C.

Answer 1

`9.38* 10^7 m/s`

Step-by-step explanation:

We are given that

Potential ,V=25 kV=
`25* 10^3 V`

Distance,r =1 cm=
`(1)/(100)=0.01 m`

1 m=100 cm

Mass of electron, m=
`9.1* 10^(-31) kg`

Charge, q=
`1.6* 10^(-19) C`

We have to find the final velocity of the electron.

Speed of electron,
`v=\sqrt{(2qV)/(m)}`

Using the formula

`v=\sqrt{(2* 1.6* 10^(-19)* 25* 10^3)/(9.1* 10^(-31))`

v=
`9.38* 10^7 m/s`

Hence, the final velocity of the electron=
`9.38* 10^7 m/s`